16 条题解

  • 4
    @ 2023-2-1 16:35:54 
    #include<iostream>
using namespace std;
int main()
{
	int a;
	cin>>a;
	for(int i=1;i<=a;i++)//第i行第一个数字为i;
	{
		for(int j=i;j<=a*a;j+=a)//第i行的数字是从i开始严格增加a，共a个，所以j<=a*a,j+=a
		{
			cout<<j<<" ";//直接输出就可以了
		}
		cout<<"\n";//注意每行后面有一个空格，（神学的说法是\n比endl快，所以这里用\n）
	}
}//完美散花
    • 3
      @ 2025-1-24 10:41:00 
      #include <bits/stdc++.h>
using namesapce std;
const int N=1e2+10;
const int INF=0x3f3f3f3f;
int n,a[N][N]; 
int main()
{
	cin<<n; 
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			a[i][j]+=n-(n-i);
			cout<<a[i][j]<<" ";
			a[i][j+1]+=n*j;
		}
		cout<<endl;
	}
return o;
}

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      • 1
        @ 2025-1-24 10:45:13 
        #include <bits/stdc++.h>
using namespace std;
const int N=1e2+10;
const int INF=0x3f3f3f3f;
int n,a[N][N];
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
a[i][j]+=n-(n-i);
cout<<a[i][j]<<" ";
a[i][j+1]+=n*j;
}
cout<<endl;
}
return 0;
}
        • 1
          @ 2024-11-10 18:19:00 
          #include<iostream>
using namespace std;
int main(){
	int N;
	cin>>N;
	int a[15][15];
	for(int i=0;i<n;i++){//第n行 
		for(int j=0;j<n;j++){//第j列 
			a[j][i]=++num;
		}
	}
	for(int i=0;i<n;j++) {
		for(int j=0;i<n;i++){
			cout<<a[i][j]<<" "; 
		}
		cout<<endl;
	}
	return 0;
}
          • 1
            @ 2024-10-25 19:41:37

            #include<bits/stdc++.h> using namespace std; #define LL long long int n[100][100]; int main() { int b , e = 1; cin >> b; for(int x = 1 ;x <= b ;x ++) { for(int v= 1; v <= b ; v++) { n[x][v] = e++; } } for(int x = 1 ; x <=b ;x++) { for(int v = 1; v <= b ; v++) cout << n[v][x] << " "; cout << endl; } }

            • 1
              @ 2024-10-25 19:28:46 
              • #include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int a[100][100];
int main()
{
	int n , len = 1;
	cin >> n;
	for(int i = 1 ;i <= n ;i ++)
	{
		for(int j= 1; j <= n ; j++)
		{
			a[i][j] = len++;
		}
	}
	for(int i = 1 ; i <=n ;i++)
	{
		for(int j = 1; j <= n ; j++)
			cout << a[j][i] << " ";
		cout << endl;
	}
	return 0;
}

                Copy

              • 1
                @ 2023-4-29 12:31:50 
                #include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int N=1e2+10;
const int INF=0x3f3f3f3f;
int n,a[N][N]; 
int main()
{
	cin>>n; 
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			a[i][j]+=n-(n-i);
			cout<<a[i][j]<<" ";
			a[i][j+1]+=n*j;
		}
		cout<<endl;
	}

}
                • 1
                  @ 2023-4-9 19:56:35 
                  #include<iostream>
using namespace std;
int main()
{
	int a;
	cin>>a;
	for(int i=1;i<=a;i++);
	{
		for(int j=i;j<=a*a;j+=a)
		{
			cout<<j<<" ";
		}
		cout<<"\n";
	}
}
                  • 0
                    @ 2025-1-24 10:44:27

                    #include <bits/stdc++.h> using namespace std; const int N=1e2+10; const int INF=0x3f3f3f3f; int n,a[N][N]; int main() { cin>>n; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { a[i][j]+=n-(n-i); cout<<a[i][j]<<" "; a[i][j+1]+=n*j; } cout<<endl; } return 0; }

                    • 0
                      @ 2024-9-14 22:30:14

                      easy

                      using namespace std;
const int N=1e2;
const int M=1e2;
const int INF=0x3f3f3f3f;
int a[N][M]={{}},n,k=1;
int main(){
cin>>n;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
a[i][j]=k;
k++;
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
cout<<a[j][i]<<" ";
}
cout<<"\n";
}
}
                      • 0
                        @ 2023-4-22 17:37:59

                        #include <bits/stdc++.h> using namespace std; int main() { int a; cin >> a; for(int i = 1; i <= a; i++) { for(int j = i; j <= a * a; j+=a) { cout << j << " "; } cout << endl; } }

                        • 0
                          @ 2023-4-22 10:07:00 
                          #include<iostream>
using namespace std;
int main()
{
    int n,m;
    cin>>n>>m;
    int arr1[n][m],arr2[n][m];
    for(int i=0;i<n;i++)for(int j=0;j<m;j++)cin>>arr1[i][j];
    for(int i=0;i<n;i++)for(int j=0;j<m;j++)cin>>arr2[i][j];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)cout<<arr1[i][j]+arr2[i][j]<<' ';
        cout<<endl;
    }
    return 0;
}
                          • 0
                            @ 2023-3-21 16:52:12 
                            #include <iostream>
using namespace std;
#define LL long long
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int a[100][100];
int main(){
	int n,len=1;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			a[i][j]=len++;
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
			cout<<a[j][i]<< " ";
		cout<<endl;
	}
	return 0;
}
                            • -1
                              @ 2024-4-9 17:20:38 
                              #include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		for(int j=i;j<=n*n;j+=n)
		{
			cout<<j<<" ";
		}
		cout<<endl;
	}
}
                              • -1
                                @ 2023-7-11 10:55:41

                                #include <stdio.h> #include using namespace std; const int N = 10+ 10; const int INF = 0x3f3f3f3f; int n, a[N][N],x,y,num; int main(){ cin >>n; x=1,y=1;//起点位置 while(num<n*n) { //向下填数 while(x<=n) a[x++][y]=++num; //回正轨 x=1,y++; } for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { cout <<a[i][j]<<" "; } cout<<endl; }

                                return 0;

                                }

                                • -1
                                  @ 2022-8-26 16:44:56 
                                  #include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int a[100][100];
int main()
{
	int n , len = 1;
	cin >> n;
	for(int i = 1 ;i <= n ;i ++)
	{
		for(int j= 1; j <= n ; j++)
		{
			a[i][j] = len++;
		}
	}
	for(int i = 1 ; i <=n ;i++)
	{
		for(int j = 1; j <= n ; j++)
			cout << a[j][i] << " ";
		cout << endl;
	}
	return 0;
}
                                  • 1

                                  二维数组输出1

                                  信息

                                  ID
                                  1042
                                  时间
                                  1000ms
                                  内存
                                  32MiB
                                  难度
                                  6
                                  标签
                                  递交数
                                  1042
                                  已通过
                                  325
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